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Proof step 7 from proof get-append-less (previous step)

n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ (a ' + x0) + y = [], ((a ' + x0) + y).tail[n - 1] = (a ' + x0)[n], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ (x0 + y)[n - 1] = (a ' + x0)[n], n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)